📚 GCSE AQA Maths: Linear Programming Revision Notes | GCSE AQA 数学：线性规划 考点精讲
Linear programming is a powerful graphical method for finding the best possible outcome – such as maximum profit or minimum cost – in a given mathematical model whose requirements are represented by linear inequalities. In the AQA GCSE Maths syllabus, it brings together skills in plotting straight lines, shading inequality regions, and interpreting real-world constraints to solve optimisation problems. This article will walk you through the key concepts, techniques, and exam tips you need to master this topic.
线性规划是一种强大的图解方法，用于在由线性不等式表示的数学模型中寻找最优结果，例如最大利润或最小成本。在 AQA GCSE 数学大纲中，它综合了绘制直线、标记不等式区域以及解读现实约束条件来解决优化问题的技能。本文将带你梳理必须掌握的核心概念、解题技巧与应试要点。
1. What is Linear Programming? | 什么是线性规划？
two-variable problems plotted on the Cartesian plane, where the region satisfying all constraints is called the feasible region.
线性规划是指在一组线性约束条件下，寻找某个线性表达式（目标函数）的最大值或最小值。所有约束条件以及目标函数都是线性的，即可以写成 ax + by + c ≤ 0、≥ 0 等形式。考试中主要出现两变量问题，在直角坐标平面上绘制，所有约束条件都满足的区域称为可行域。
2. Linear Inequalities and Their Graphs | 线性不等式及其图形
nd a dashed line for < or >. Then shade the side of the line that satisfies the inequality – test a point, usually (0,0), unless the line passes through the origin.
在解决线性规划问题之前，你必须能够熟练地将线性不等式转化为图形。以 y ≤ 2x + 1 为例，先画出边界线 y = 2x + 1。对于 ≤ 或 ≥ 使用实线，对于 < 或 > 使用虚线。然后对满足不等式的那一侧进行涂色——除非直线经过原点，否则通常选取 (0,0) 作为测试点。
line y = 3x - 2 is dashed. Substitute (0,0): 0 < -2 is false, so shade the opposite side.
示例：对于 y < 3x - 2，直线 y = 3x - 2 为虚线。代入 (0,0)：0 < -2 为假，因此涂色的区域是另一侧。
Multiple inequalities create a system. The solution is the intersection (overlap) of all shaded regions.
多个不等式构成一个系统，其解集是所有涂色区域的交集（重叠部分）。
3. Representing Inequality Regions | 表示不等式区域
In an exam, you might be given a graph with several lines and asked to write the inequalities that define a shaded region. Identify each line's equation, determine whether the line is solid or dashed, and then test a point inside the region to decide the correct inequality sign. Always state the inequalities in their simplest integer form where possible.
考试中可能会给你一幅带有若干条直线的图形，要求你写出定义涂色区域的不等式。先识别每条直线的方程，判断直线为实线还是虚线，然后在区域内取一个测试点以确定正确的不等号方向。可能的话，始终将不等式写成最简整数形式。
l and horizontal lines: x ≥ a, x ≤ a, y ≥ b, y ≤ b. Make sure to use the correct variable.
对于垂直和平行于坐标轴的直线：x ≥ a, x ≤ a, y ≥ b, y ≤ b。务必使用正确的变量。
4. The Feasible Region | 可行域
quadrant. The feasible region is always a polygon (often a quadrilateral or triangle) whose vertices are the candidate points for optimal solutions.
可行域是同时满足所有约束条件的点的集合。在线性规划问题中，通常还要求 x ≥ 0 且 y ≥ 0，意味着我们在第一象限内研究。可行域总是一个多边形（通常为四边形或三角形），其顶点就是最优解的候选点。
To find the feasible region: graph each constraint, shade the unwanted region for each (the side that does NOT satisfy the inequality), and the unshaded region left is your feasible region. Many exam boards prefer shading out, but AQA often shades the feasible region itself – watch the wording.
确定可行域的方法：画出每个约束条件，对每个条件将不满足的区域涂掉（即涂掉不满足不等式的那一侧），最后留下的未涂区域就是可行域。很多考试局习惯将不可行区域涂色，但 AQA 常直接涂出可行域——关注题目用词。
5. The Objective Function | 目标函数
does not affect the feasible region but is used to evaluate the vertices to find the best value. Write it clearly and be ready to substitute coordinates.
目标函数是你试图最大化或最小化的线性表达式，例如利润 P = 3x + 2y 或成本 C = 5x + 4y。该函数不会影响可行域，但需要用顶点的坐标代入求值，以找到最优解。清晰地写出目标函数，并准备好代入坐标。
In some problems, you may need to construct the objective function from words: 'profit is £5 per cake and £3 per bun' gives P = 5x + 3y. Ensure you define x and y clearly from the start.
有些问题需要你从文字中构造目标函数：例如“每个蛋糕利润 £5，每个面包利润 £3”给出 P = 5x + 3y。务必从一开始就明确定义 x 和 y 的含义。
6. The Vertex Method for Maximising Profit | 顶点法最大化利润
For GCSE, the method is simple: once the feasible region is drawn, identify the coordinates of all vertices (corner points). This often involves solving simultaneous equations for the intersection of two boundary lines. Substitute each vertex into the objective function and pick the vertex giving the highest value (for maximisation) or lowest (for minimisation).
对于 GCSE 而言，方法很简单：画出可行域后，确定所有顶点（角点）的坐标。这通常需要解两条边界线交点的联立方程。将每个顶点代入目标函数，选择给出最大值（最大化问题）或最小值（最小化问题）的顶点即可。
(5,0). Evaluate: P(0,0)=0, P(0,4)=12, P(4,2)=22, P(5,0)=20. Maximum profit is 22 at (4,2).
例题：在约束条件 x + 2y ≤ 8、2x + y ≤ 10、x ≥ 0、y ≥ 0 下最大化 P = 4x + 3y。顶点：(0,0)、(0,4)、(4,2)、(5,0)。代入：P(0,0)=0，P(0,4)=12，P(4,2)=22，P(5,0)=20。最大利润为 22，位于 (4,2)。
7. Minimisation Problems | 最小化问题
Linear programming can also be used to minimise costs or materials. The process is identical except you look for the smallest value of the objective function at the vertices. Pay attention to non-negativity constraints: occasionally the minimum might occur on an axis, but the vertex method still holds because the minimum of a linear function over a convex polygon occurs at a vertex.
线性规划也可用于最小化成本或材料用量。除了要在顶点处寻找目标函数的最小值外，求解过程完全相同。注意非负约束：最小值有时可能出现在坐标轴上，但顶点法仍然适用，因为线性函数在凸多边形上的最小值必然出现在顶点处。
If the objective function is parallel to one side of the feasible region, there may be infinitely many solutions along that edge. At GCSE level, any vertex on that edge is acceptable.
如果目标函数与可行域的某一条边平行，那么该边上的所有点都是最优解，此时有无数个解。在 GCSE 层面，选取该边上的任一顶点作为答案即可。
8. Real-World Applications | 实际应用题
ls, labour). Define your decision variables (let x be the number of type A, y the number of type B). Translate sentences such as 'each A takes 2 hours, each B takes 3 hours, total time cannot exceed 50 hours' into 2x + 3y ≤ 50.
典型的 GCSE 题目会设置一个场景：工厂生产两种产品。约束条件来自有限的资源（机器工时、原材料、劳动力）。定义决策变量（设 x 为 A 类产品数量，y 为 B 类产品数量）。将“每个 A 需要 2 小时，每个 B 需要 3 小时，总时间不超过 50 小时”这样的语句转化为 2x + 3y ≤ 50。
You may also need to integer constraints: 'x and y must be whole numbers'. Though the graph uses continuous regions, the optimal solution must be an integer point inside or on the feasible region. Check integers near the continuous optimum.
还可能涉及整数约束：“x 和 y 必须为整数”。虽然图形使用连续区域，但最优解必须是可行域内部或边界上的整数点。需要检查连续最优解附近的整数点。
9. Common Mistakes and Tips | 常见错误与提示
Forgetting to draw lines accurately: use a ruler, label axes, and plot at least two points for each line. A sloppy graph can lead to wrong vertex coordinates.
忘记精确作图：使用直尺，标注坐标轴，每条直线至少描两个点。草率的图形会导致错误的顶点坐标。
Mixing up shading direction: always test a point. If the inequality involves 'y ≥ ...', shade above the line; 'y ≤ ...' shade below – but only if y is isolated. Better to rely on point testing.
搞错涂色方向：始终测试一个点。如果不等式是“y ≥ …”的形式，则涂色在线条上方；“y ≤ …”涂色在下方——但只有将 y 单独表示时这条规则才成立。更可靠的方法是依靠测试点。
Missing the non-negativity constraints: unless told otherwise, assume x ≥ 0, y ≥ 0.
遗漏非负约束：除非题目另有说明，否则默认 x ≥ 0, y ≥ 0。
Misidentifying vertices: only consider intersections that lie within all constraints, not all intersections of lines.
错误识别顶点：只考虑位于所有约束条件内的交点，而不是所有直线的交点。
Forgetting to answer the question: after finding (x,y), state clearly 'maximum profit is £... when producing ... of A and ... of B'.
忘记回答题目所问：在找到 (x, y) 后，要清楚地陈述“最大利润为 £…，此时生产 A 产品 … 件，B 产品 … 件”。
10. Constructing Inequalities from Diagrams | 根据图形列写不等式
You might see a shaded triangle or quadrilateral on a grid. The question asks: 'Write down the three inequalities that define this region.' If (0,1) is inside, for x we need x ≥ -1; for y we need y ≤ 2; and for y = 2x we check: 1 ? o inequality is y > 2x (or y ≥ 2x if line is solid). Be careful with strict or non-strict.
你可能会看到网格上的一个涂色三角形或四边形。题目要求：“写出定义该区域的三条不等式”。找出各边的方程。例如，由 x = -1、y = 2 以及经过 (0,0) 和 (2,4) 的直线围成的三角形，其斜边方程为 y = 2x。然后通过在三角形内取点（如 (0,1)）判断不等号方向。如果 (0,1) 在内部，对于 x 我们需要 x ≥ -1；对于 y 需要 y ≤ 2；对于 y = 2x 检验：1 ? 2(0) → 1 > 0，因此不等式为 y > 2x（若直线为实线则用 y ≥ 2x）。注意严格与非严格不等号的区别。
11. Exam-Style Example Walkthrough | 考试风格例题讲解
Problem: A small business makes two types of gift hamper: Standard (x) and Luxury (y). Each Standard hamper requires 2 hours of preparation and 1 hour of packing. Each Luxury hamper requires 1 hour of preparation and 3 hours of packing. In a week, the business has a maximum of 40 preparation hours and 45 packing hours. The profit is £15 per Standard hamper and £25 per Luxury hamper. Find the maximum weekly profit.
题目：一家小企业生产两种礼篮：标准版 (x) 和豪华版 (y)。每个标准礼篮需要 2 小时准备时间和 1 小时包装时间。每个豪华礼篮需要 1 小时准备时间和 3 小时包装时间。每周企业最多有 40 小时准备时间和 45 小时包装时间。每个标准礼篮利润 £15，每个豪华礼篮利润 £25。求每周最大利润。
Step 1 – Define variables and constraints: x ≥ 0, y ≥ 0 (non-negativity). Preparation: 2x + y ≤ 40. Packing: x + 3y ≤ 45.
第 1 步 – 定义变量和约束条件：x ≥ 0, y ≥ 0（非负）。准备时间：2x + y ≤ 40。包装时间：x + 3y ≤ 45。
Step 2 – Graph and find vertices: 45 → -5x = -75 → x = 15, then y = 40 - 30 = 10. Vertices: (0,0), (0,15), (15,10), (20,0).
第 2 步 – 作图并求顶点：画出 2x + y = 40（点：(0,40), (20,0)）。画出 x + 3y = 45（点：(0,15), (45,0)）。交点：联立求解。由 2x + y = 40 得 y = 40 - 2x。代入：x + 3(40 - 2x) = 45 → x + 120 - 6x = 45 → -5x = -75 → x = 15，然后 y = 40 - 30 = 10。顶点：(0,0), (0,15), (15,10), (20,0)。
Step 3 – Evaluate objective function P = 15x + 25y: P(0,0)=0, P(0,15)=375, P(15,10)=225+250=475, P(20,0)=300. Maximum £475 at (15,10).
第 3 步 – 代入目标函数 P = 15x + 25y：P(0,0)=0, P(0,15)=375, P(15,10)=225+250=475, P(20,0)=300。最大值 £475，位于 (15,10)。
Answer: Produce 15 Standard and 10 Luxury hampers for a maximum profit of £475 per week.
答案：每周生产 15 个标准礼篮和 10 个豪华礼篮，可获得最大利润 £475。
12. Key Summary for the Exam | 考试要点总结
Always define your variables at the start.
始终从定义变量开始。
Convert each constraint into a linear inequality and graph accurately.
将每个约束条件转化为线性不等式并精确作图。
Find the feasible region where all shaded areas overlap.
找出所有涂色区域重叠的可行域。
Identify vertices by solving pairs of boundary equations.
通过解边界方程对来识别顶点。
Substitute each vertex into the objective function; choose the best value.
将每个顶点代入目标函数；选择最优值。
Interpret your solution in the context of the problem, with correct units.
在题目情境中解释你的解，并带上正确单位。
If integer answers are needed, check points around the continuous optimum.
如果需要整数解，检查连续最优解附近的点。
Practice past paper questions to build speed and confidence.
通过练习历年真题提高速度和信心。
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